Integrand size = 24, antiderivative size = 109 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {8 i (a-i a \tan (c+d x))^7}{7 a^{10} d}-\frac {3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac {2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac {i (a-i a \tan (c+d x))^{10}}{10 a^{13} d} \] Output:
8/7*I*(a-I*a*tan(d*x+c))^7/a^10/d-3/2*I*(a-I*a*tan(d*x+c))^8/a^11/d+2/3*I* (a-I*a*tan(d*x+c))^9/a^12/d-1/10*I*(a-I*a*tan(d*x+c))^10/a^13/d
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.51 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {(i+\tan (c+d x))^7 \left (-44-98 i \tan (c+d x)+77 \tan ^2(c+d x)+21 i \tan ^3(c+d x)\right )}{210 a^3 d} \] Input:
Integrate[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^3,x]
Output:
((I + Tan[c + d*x])^7*(-44 - (98*I)*Tan[c + d*x] + 77*Tan[c + d*x]^2 + (21 *I)*Tan[c + d*x]^3))/(210*a^3*d)
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{14}}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^6 (i \tan (c+d x) a+a)^3d(i a \tan (c+d x))}{a^{13} d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left (-(a-i a \tan (c+d x))^9+6 a (a-i a \tan (c+d x))^8-12 a^2 (a-i a \tan (c+d x))^7+8 a^3 (a-i a \tan (c+d x))^6\right )d(i a \tan (c+d x))}{a^{13} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {8}{7} a^3 (a-i a \tan (c+d x))^7+\frac {3}{2} a^2 (a-i a \tan (c+d x))^8+\frac {1}{10} (a-i a \tan (c+d x))^{10}-\frac {2}{3} a (a-i a \tan (c+d x))^9\right )}{a^{13} d}\) |
Input:
Int[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*((-8*a^3*(a - I*a*Tan[c + d*x])^7)/7 + (3*a^2*(a - I*a*Tan[c + d*x]) ^8)/2 - (2*a*(a - I*a*Tan[c + d*x])^9)/3 + (a - I*a*Tan[c + d*x])^10/10))/ (a^13*d)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 0.54 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.53
method | result | size |
risch | \(\frac {128 i \left (120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\) | \(58\) |
derivativedivides | \(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )^{10}}{10}-\frac {i \tan \left (d x +c \right )^{9}}{3}-\frac {8 i \tan \left (d x +c \right )^{7}}{7}+\tan \left (d x +c \right )^{6}-\frac {6 i \tan \left (d x +c \right )^{5}}{5}+2 \tan \left (d x +c \right )^{4}+\frac {3 \tan \left (d x +c \right )^{2}}{2}\right )}{a^{3} d}\) | \(91\) |
default | \(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )^{10}}{10}-\frac {i \tan \left (d x +c \right )^{9}}{3}-\frac {8 i \tan \left (d x +c \right )^{7}}{7}+\tan \left (d x +c \right )^{6}-\frac {6 i \tan \left (d x +c \right )^{5}}{5}+2 \tan \left (d x +c \right )^{4}+\frac {3 \tan \left (d x +c \right )^{2}}{2}\right )}{a^{3} d}\) | \(91\) |
Input:
int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
128/105*I*(120*exp(6*I*(d*x+c))+45*exp(4*I*(d*x+c))+10*exp(2*I*(d*x+c))+1) /d/a^3/(exp(2*I*(d*x+c))+1)^10
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (85) = 170\).
Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.78 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {128 \, {\left (-120 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{105 \, {\left (a^{3} d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, a^{3} d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-128/105*(-120*I*e^(6*I*d*x + 6*I*c) - 45*I*e^(4*I*d*x + 4*I*c) - 10*I*e^( 2*I*d*x + 2*I*c) - I)/(a^3*d*e^(20*I*d*x + 20*I*c) + 10*a^3*d*e^(18*I*d*x + 18*I*c) + 45*a^3*d*e^(16*I*d*x + 16*I*c) + 120*a^3*d*e^(14*I*d*x + 14*I* c) + 210*a^3*d*e^(12*I*d*x + 12*I*c) + 252*a^3*d*e^(10*I*d*x + 10*I*c) + 2 10*a^3*d*e^(8*I*d*x + 8*I*c) + 120*a^3*d*e^(6*I*d*x + 6*I*c) + 45*a^3*d*e^ (4*I*d*x + 4*I*c) + 10*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)
\[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{14}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**14/(a+I*a*tan(d*x+c))**3,x)
Output:
I*Integral(sec(c + d*x)**14/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan (c + d*x) + I), x)/a**3
Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{10} + 70 \, \tan \left (d x + c\right )^{9} + 240 \, \tan \left (d x + c\right )^{7} + 210 i \, \tan \left (d x + c\right )^{6} + 252 \, \tan \left (d x + c\right )^{5} + 420 i \, \tan \left (d x + c\right )^{4} + 315 i \, \tan \left (d x + c\right )^{2} - 210 \, \tan \left (d x + c\right )}{210 \, a^{3} d} \] Input:
integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/210*(-21*I*tan(d*x + c)^10 + 70*tan(d*x + c)^9 + 240*tan(d*x + c)^7 + 2 10*I*tan(d*x + c)^6 + 252*tan(d*x + c)^5 + 420*I*tan(d*x + c)^4 + 315*I*ta n(d*x + c)^2 - 210*tan(d*x + c))/(a^3*d)
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{10} + 70 \, \tan \left (d x + c\right )^{9} + 240 \, \tan \left (d x + c\right )^{7} + 210 i \, \tan \left (d x + c\right )^{6} + 252 \, \tan \left (d x + c\right )^{5} + 420 i \, \tan \left (d x + c\right )^{4} + 315 i \, \tan \left (d x + c\right )^{2} - 210 \, \tan \left (d x + c\right )}{210 \, a^{3} d} \] Input:
integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-1/210*(-21*I*tan(d*x + c)^10 + 70*tan(d*x + c)^9 + 240*tan(d*x + c)^7 + 2 10*I*tan(d*x + c)^6 + 252*tan(d*x + c)^5 + 420*I*tan(d*x + c)^4 + 315*I*ta n(d*x + c)^2 - 210*tan(d*x + c))/(a^3*d)
Time = 0.64 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\cos \left (c+d\,x\right )}^{10}\,84{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^9+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+40\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3-{\cos \left (c+d\,x\right )}^2\,105{}\mathrm {i}-70\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )+21{}\mathrm {i}}{210\,a^3\,d\,{\cos \left (c+d\,x\right )}^{10}} \] Input:
int(1/(cos(c + d*x)^14*(a + a*tan(c + d*x)*1i)^3),x)
Output:
(40*cos(c + d*x)^3*sin(c + d*x) - 70*cos(c + d*x)*sin(c + d*x) + 48*cos(c + d*x)^5*sin(c + d*x) + 64*cos(c + d*x)^7*sin(c + d*x) + 128*cos(c + d*x)^ 9*sin(c + d*x) - cos(c + d*x)^2*105i + cos(c + d*x)^10*84i + 21i)/(210*a^3 *d*cos(c + d*x)^10)
Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.81 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9}+576 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}-1008 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-210 \cos \left (d x +c \right ) \sin \left (d x +c \right )-42 \sin \left (d x +c \right )^{10} i +210 \sin \left (d x +c \right )^{8} i -420 \sin \left (d x +c \right )^{6} i +420 \sin \left (d x +c \right )^{4} i -315 \sin \left (d x +c \right )^{2} i +126 i}{210 a^{3} d \left (\sin \left (d x +c \right )^{10}-5 \sin \left (d x +c \right )^{8}+10 \sin \left (d x +c \right )^{6}-10 \sin \left (d x +c \right )^{4}+5 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x)
Output:
( - 128*cos(c + d*x)*sin(c + d*x)**9 + 576*cos(c + d*x)*sin(c + d*x)**7 - 1008*cos(c + d*x)*sin(c + d*x)**5 + 840*cos(c + d*x)*sin(c + d*x)**3 - 210 *cos(c + d*x)*sin(c + d*x) - 42*sin(c + d*x)**10*i + 210*sin(c + d*x)**8*i - 420*sin(c + d*x)**6*i + 420*sin(c + d*x)**4*i - 315*sin(c + d*x)**2*i + 126*i)/(210*a**3*d*(sin(c + d*x)**10 - 5*sin(c + d*x)**8 + 10*sin(c + d*x )**6 - 10*sin(c + d*x)**4 + 5*sin(c + d*x)**2 - 1))